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Dependent on what side you are given you would use Sin(Θ) = Opposite/Hypotenuse just rearrange the formula to Hypotenuse = Opposite/Sin(Θ). Or if you are given the adjacent side use Cosine(Θ)=Adjacent/Hypotenuse, then: Hypotenuse = Adjacent/Cosine(Θ)
The basic functions are sine, cosine, tangent, cosecant, secant and cotangent. In addition, there are their inverses, whose full names use the prefix "arc" [arcsine, arc cosine, etc] but are more often written as sin-1, cos-1 and so on.
There are two ways to solve for the double angle formulas in trigonometry. The first is to use the angle addition formulas for sine and cosine. * sin(a + b) = sin(a)cos(b) + cos(a)sin(b) * cos(a + b) = cos(a)cos(b) - sin(a)sin(b) if a = b, then * sin(2a) = sin(a)cos(a) + cos(a)sin(a) = 2sin(a)cos(a) * cos(2a) = cos2(a) - sin2(b) The cooler way to solve for the double angle formulas is to use Euler's identity. eix = cos(x) + i*sin(x). Yes, that is "i" as in imaginary number. we we put 2x in for x, we get * e2ix = cos(2x) + i*sin(2x) This is the same as * (eix)2 = cos(2x) + i*sin(2x) We can substitute our original equation back in for eix. * (cos(x) + i*sin(x))2 = cos(2x) + i*sin(2x) We can distribute the squared term. * cos2(x) + i*sin(x)cos(x) + i*sin(x)cos(x) + (i*sin(x))2 = cos(2x) + i*sin(2x) And simplify. Because i is SQRT(-1), the i squared term becomes negative. * cos2(x) + 2i*sin(x)cos(x) - sin2(x) = cos(2x) + i*sin(2x) * cos2(x) - sin2(x) + 2i*sin(x)cos(x) = cos(2x) + i*sin(2x) Now you can plainly see both formulas in the equation arranged quite nicely. I don't yet know how to get rid of the i, but I'm working on it.
you use the the 3 trigonometry functions , sin=opposite divided by hypotenuse cos=adjacent divided by hypotenuse tan=opposite divided by adjacent these are used to work out angles and side lengths in right angle triangles only!!! sine,cosine,tangent :)
It depends on what else you know. If it is a non-right triangle, and you only know angle a, it is impossible to fing side A (the side opposite an angle usually has the same letter, but capitalized). If you know the other two sides, then I would use the law of cosines: For a triangle with sides A B C A = √(B2+C2-(2*B*C*(cos (a)))) If you know another angle and one side, I would use the law sines: A/(sin a) = B/(sin b) therefore, A = (sin a) * B/(sin b) If it is a right triangle, and you know another side, than your job is even easier: If you know the hypotenuse (side C), than: A = C *(sin a) If you know the adjacent side (side B), than: A = B * (tan a)
You can choose either or but tangent which is sin/cos seems to be the most common way.
Dependent on what side you are given you would use Sin(Θ) = Opposite/Hypotenuse just rearrange the formula to Hypotenuse = Opposite/Sin(Θ). Or if you are given the adjacent side use Cosine(Θ)=Adjacent/Hypotenuse, then: Hypotenuse = Adjacent/Cosine(Θ)
It depends on whether you know the lengths of all three sides (either explicitly or otherwise). If you don't know the lengths of the sides you cannot find the top angle. If you do know the sides you can apply the cosine rule: cos(A) = (b2 +c2 - a2)/2bc and then use the inverse cosine function to determine A.
The inverse (negatives) of sine, cosine, and tangent are used to calculate the angle theta (or whatever you choose to name it). Initially it is taught that opposite over hypotenuse is equal to the sine of theta sin(theta) = opposite/hypotenuse So it can be said that theta = sin-1 (opp/hyp) This works the same way with cosine and tangent In short the inverse is simply what you use when you move the sin, cos, or tan to the other side of the equation generally to find the angle
Two methods to try . #1 Use pythagoras h^ = a^2 + a^2 NB THis is only good if you know that the two unknown sides are the same length. #2 Use trigonometry (trig.) This is good if you know the hypotenuse and one of the angles. Sine(angle) = opposite/ hypotenuse Hence opposite side = hypotenuse X sine(angle) Similarly Cosine(angle) = adjacent / hypotenuse. adjacent side = hypotenuse X Cosine(angle) Here is an example If you known the hypotenuse is a length of '6' and the angle is 30 degrees. Then opposite = 6 X Sin(30) opposite = 6 x 0.5 = 3 So the length of the oppisute sides is '3' units. NB DO NOT make the mistakes of saying Sin(6 X 30) = Sin(180) Nor 6 x 30 , nor Sin(6) X 30 , nor any other combination. You MUST find the SINE of the angle , then multiply it to the given length. Similarly for Cosine and Tangent.
pen0r
The sine and cosine were originally developed for use in surveying. They provided a way to measure the distance across lakes and around mountains. Soon they were found to be useful in navigation. The sine was used to calculate pi. When electrical measurements were made, the sine law was used. If you want to know when to use the sine and when to use the cosine, you will need to get a trig book, a physics book, an astronomy book, a sailing book, and a few other books and read them all.
53.5 square cm to 1 d.p. Use the cosine rule to find an angle. Then use: Area = 1/2*a*b*sin C
By its very mane, a sinusoidal wave refers to a sine function. The cosine function is simply the sine function that is phase-shifted.
cosine = adjacent/hypotenuse. It can be used as other trig functions can.
If you do not know only a side length you cannot. If you know all three side lengths then you can use the cosine rule. You can continue using the cosine rule for the other two angles but, once you have one angle, it is simpler to use the sine rule.
The basic functions are sine, cosine, tangent, cosecant, secant and cotangent. In addition, there are their inverses, whose full names use the prefix "arc" [arcsine, arc cosine, etc] but are more often written as sin-1, cos-1 and so on.